Velocity and Acceleration

Main idea

If $\vec r(t)$ is where something is, then $\vec r\,'$ is its velocity, the length of that is its speed, and $\vec r\,''$ is its acceleration. Acceleration does two jobs, and it splits neatly into a part that changes your speed and a part that changes your direction.

\vec v(t) = \vec r\,'(t), \qquad \vec a(t) = \vec r\,''(t), \qquad \text{speed} = \|\vec v(t)\|

Tangential and normal components

\vec a = {\htmlData{sym=0}{a_T}}\,\vec T + {\htmlData{sym=1}{a_N}}\,\vec N, \qquad {\htmlData{sym=0}{a_T}} = \frac{\vec v\cdot\vec a}{\|\vec v\|}, \qquad {\htmlData{sym=1}{a_N}} = \frac{\|\vec v\times\vec a\|}{\|\vec v\|} = \kappa\|\vec v\|^2

What the symbols mean

$a_T$
Tangential acceleration
Speeding up or slowing down.
$a_N$
Normal acceleration
Turning: the centripetal part, κ times speed squared.

Explanation

Velocity is simply how fast you're going and which way, and acceleration is how that velocity is changing. The key surprise is that you can accelerate without speeding up: just turning the steering wheel counts. The split $\vec a = a_T\,\vec T + a_N\,\vec N$ looks technical, but it's two plain pieces: $a_T$ (the gas or brake pedal, changing your speed along the road) and $a_N$ (the steering wheel, pulling you sideways into the turn). Every acceleration is some mix of those two. That's all the formulas are saying.

We've already done most of the work. The derivative $\vec r\,'$ points along the path with length equal to the speed, so it's the velocity. Differentiate once more and you get the acceleration, $\vec r\,''$ , just like position, velocity and acceleration in one-variable calculus, but now each one is an arrow.

Splitting acceleration. Why does it split into exactly $\vec T$ and $\vec N$ ? Write $\vec v = \|\vec v\|\,\vec T$ (speed times direction) and use the product rule: the derivative of the speed gives a piece along $\vec T$ , and the derivative of $\vec T$ gives a piece along $\vec N$ , by the T, N, B lesson. That second piece works out to $\kappa\|\vec v\|^2$ , so turning a sharper bend, or taking the same bend faster, takes more sideways push. Notice it's speed squared: double your speed through a turn and you need four times the grip. To compute the parts, $a_T$ is the shadow of $\vec a$ on $\vec v$ (a dot product), and $a_N$ comes from a cross product.

Projectiles. With gravity alone, $\vec a = \langle 0, -g\rangle$ . Run the process backward: integrate once from the launch velocity $\langle v_0\cos\theta, v_0\sin\theta\rangle$ to get the velocity, then once more from the launch point at the origin to get

\vec r(t) = \left\langle v_0 t\cos\theta,\; v_0 t\sin\theta - \tfrac12 gt^2\right\rangle.

Notice the horizontal part never changes speed, because gravity only pulls down.

Circular motion at constant speed has no tangential acceleration, since the speed isn't changing. All that's left is $a_N = \frac{v^2}{R}$ , pointing toward the center (on a circle $\kappa = \frac1R$ ). Something has to supply that inward pull, like the friction of your tires or the seat of the car. That's why you feel pushed outward in a turn even though your speed doesn't change: your body wants to go straight, and the car is pulling it around.

See it

A ball launched with velocity ⟨12, 15⟩ m/s, landing about 37 m away after 3.06 s. The velocity keeps turning to follow the path, while the acceleration never changes: gravity, 9.8 m/s² straight down. Notice the speed starts at about 19.2 m/s, bottoms out at 12 m/s at the top of the arc, where the ball moves purely sideways, and climbs back to 19.2 on the way down.

Worked examples

Example 1

An object moves on $\vec r(t) = \langle R\cos\omega t, R\sin\omega t\rangle$ . Find its acceleration.

1
Differentiate once, using the chain rule: $\vec v = R\omega\langle -\sin\omega t, \cos\omega t\rangle$ , so the speed is a constant $R\omega$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

For $\vec r(t) = \langle t^2, 2t, t^3\rangle$ , find $\vec v(1)$ .

Problem 2+5 XP

Find $\vec a(1)$ .

Problem 3+5 XP

Find the speed at $t = 1$ .

Problem 4+5 XP

An object moves in a circle of radius 2 with $\omega = 3$ rad/s. Find the magnitude of its acceleration.

Problem 5+5 XP

For $\vec r(t) = \langle t, t^2\rangle$ , find the tangential acceleration $a_T$ at $t = 1$ .

Problem 6+5 XP

A car speeds up at $2$ m/s² while rounding a bend of radius $40$ m, at a moment when its speed is $10$ m/s. How big is its total acceleration?

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

A ball is launched from the ground at $25$ m/s toward a target $50$ m away and $10$ m up. Two launch angles hit it. Find the lower one, in degrees. (Use $g = 9.8$ m/s².)

Where it's used

Engineers decompose acceleration this way to design safe roads, comfortable trains and the g-forces on roller coasters and fighter pilots.

Projectile MotionAlgebra · Physics

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